Báo cáo toán học: "Hurwitz Equivalence in Dihedral Groups" pptx

When the tuple T ∈ Gn consists only of reflections, the orbits are determined by the following invariants: the product of the entries, the subgroup generated by the entries, and the numb

Hurwitz Equivalence in Dihedral Groups

Emily Berger

Massachusetts Institute of Technology

ERB90@mit.edu Submitted: Nov 18, 2009; Accepted: Jan 25, 2010; Published: Feb 21, 2011

Mathematics Subject Classification: 20F36, 20F55

Abstract

In this paper we determine the orbits of the braid group Bnaction on Gnwhen

G is a dihedral group and for any T ∈ Gn We prove that the following invariants serve as necessary and sufficient conditions for Hurwitz equivalence They are: the product of its entries, the subgroup generated by its entries, and the number of times each conjugacy class (in the subgroup generated by its entries) is represented

in T

Introduction

is a dihedral group When the tuple T ∈ Gn consists only of reflections, the orbits are determined by the following invariants: the product of the entries, the subgroup generated

by the entries, and the number of times each conjugacy class (in the subgroup generated

by its entries) is represented in T

Our study of Hurwitz equivalence in the dihedral group was inspired by the paper [1], which gives a simple criterion for Hurwitz equivalence in the symmetric group analogous

to our Main Theorem That paper studies tuples of transpositions in the symmetric group, which is the reason why we originally chose to restrict to reflections in the dihedral group (Recall that the symmetric group Sm acts on Rm−1 in such a way that every transposition acts by a Euclidean reflection.) Utlimately, we extend these results to include rotations

as well

After the bulk of this work was completed we discovered the paper [3] that considers, using a different method, the case of a dihedral group of order 2pα where p is prime Our results were obtained independently and cover the case of dihedral groups of any order In addition, after this paper was finished, [5] was published, extending the results of [3] The results of our paper are complementary to the work in [5], since our results are derived from first principles using what is perhaps a more intuitive approach

1 Definitions

The braid group on n strands, Bn, may be described by n − 1 generators σ1, , σn−1 and the following defining relations

σiσj = σjσi if |i − j| ≥ 2

σiσi+1σi = σi+1σiσi+1

Consider Gn, the set of tuples of length n with entries in G The braid group acts on Gn

by Hurwitz moves Let T = (a1, a2, , an) with ai ∈ G In this sense, σi, a Hurwitz move, may be realized as the following

σiT = (a1, , aiai+1a−1i, ai, , an)

Clearly, σi and σj commute when |j − i| ≥ 2 The second relation is more subtle Assume

T has length three for simplicity

σ1σ2σ1T =  a1a2a−11
a1a3a−11
a1a2a−11
−1, a1a2a−11, a1

= a1a2a3a−12 a−11, a1a2a−11, a1

= σ2σ1σ2T Also, inverse Hurwitz moves are defined by σ−1i ( ai, ai+1, ) → ( ai+1, a−1i+1aiai+1 ) With this action, we may study the orbits of the elements of Gn, motivating the following definition

contained in the same orbit

2 Necessary Conditions for Hurwitz Equivalence

Let T = (a1, , an) and T0 = (a01, , a0n) be elements of Gn Certain properties of T are invariant under Hurwitz moves These properties will serve as necessary conditions for Hurwitz Equivalence

2.1 Product of the Elements T

DefineQ T = a1a2 an, then T ∼ T0 impliesQ T = Q T0

Proof Any σi transforms T = ( , ai, ai+1, ) to ˜T = ( , aiai+1a−1i, ai, )

Y ˜T = a1 aiai+1a−1

i ai an = a1 aiai+1 an=YT

Suppose T and T0 generate subgroups S and S0 respectively, if T ∼ T0 then S = S0

T0 If a and b are in S, so is aba−1, so S ⊆ S0 By symmetry and the use of inverse Hurwitz moves, S0 ⊆ S, so S = S0

S = S0 appears in T is the same as in T0

Proof Notice that σi acts as the transposition (i i+1) on conjugacy classes in T Without loss of generality, let i = 1 and n = 2

σ1(a1, a2) = (a1a2a−11, a1) Clearly, a1 is in the conjugacy class of a1 and a1a2a−11 in that of a2 Therefore, σi only transposes elements of conjugacy classes, and thus leaves the number of elements in each conjugacy class fixed

entries are elements of Dm The necessary conditions stated above for an arbitrary group

G serve as sufficient conditions for T ∼ T0

We first prove the main theorem for T containing only reflections, we call this the reflection main theorem We then generalize to all T ∈ Dmn

3 Preliminaries and the Main Lemma

Before proving the reflection main theorem, we fix notation and present elementary facts about the dihedral group In addition, we prove the main lemma which will be used in Section 4

3.1 Notation

We define notation by labeling the vertices and edges of a polygon Firstly, alter the polygon by adjoining a vertex to the mid-point of each edge Begin by labeling some adjoined vertex 1 and continue in the counterclockwise direction alternately numbering adjoined vertices and regular vertices 1 through m twice Images of the numbering for

m = 5 and m = 6 are below

Define the line connecting the pair of vertices (adjoined or normal) labeled i to be li and the reflection fixing li to be rli, or simply ri In addition, define the distance between two reflections d(ri, rj) to be the length of the minimal path through adjoined and regular vertices connecting some vertex on li to some vertex on lj

Figure 1: Numbering of reflections

In order to understand the action of Bn, conjugation of reflections by reflections and products of reflections must be explained

In general, conjugation by a reflection has the following formula

rirjri = rri(lj) where ri(lj) represents the line to which ri maps lj Geometrically, ri(lj) is the line symmetric to lj with respect to reflecting about li, namely lk where k − i = i − j or

k = i + (i − j)

Lemma 3.1

rirjri = ri+(i−j) Corollary 3.2 The product rirjri may also be written as rj+2(i−j) which shows that when

m is even, not all reflections are conjugate to each other They are split into edge-edge refections and vertex-vertex reflections because rk and rk0 are conjugate if and only if

k0− k ≡ 0 (mod 2)

3.2.2 Product of two reflections

Consider the product of two reflections, say rirj The product of any two reflections must

be some rotation By definition, rj fixes lj, so the rotation is determined by which line lj gets mapped to by ri Geometrically, it is clear that this line is lk where i − j = k − i Therefore, if we fix counterclockwise to be the positive direction, rirj is a rotation through (i − j)2π

m

Lemma 3.3

The product rirj is a rotation through (i − j)2π

m. Lemma 3.4 The the orbit of (ri, rj) is O = {(ri+k(i−j), ri+(k−1)(i−j)) | k ∈ Zm}

Proof By Lemma 3.1,

(ri, rj) ∼ σ(ri, rj) = (rirjri, ri) = (ri+(i−j), ri)

Since i + (i − j) − i = i − j, the above shows σi does not change the the difference between the ith and i + 1st entries For fixed k we have

σ(ri+k(i−j), ri+(k−1)(i−j)) = (ri+(k+1)(i−j), ri+(k)(i−j)) since

i + (k + 1)(i − j) = i + k(i − j) +i + k(i − j)−i + (k − 1)(i − j)

We apply σ (at times we will omit the i attached to σi) in repetition to obtain the orbit

O of (ri, rj)

O = {(ri+k(i−j), ri+(k−1)(i−j)) | k ∈ Zm}

We remark that the size of the orbit is determined by the smallest k > 0 such that k(i − j) ≡ 0 (mod m) At this time, the first entry of the pair has returned to ri, causing the second to return to rj

divides m

D Corollary 3.6 If the gcd(i − j, m) = 1, the orbit of (ri, rj) is of size m and contains all pairs (ri0, rj0) where i0 − j0 = i − j In otherwords, (rk, rk−(i−j)) ∈ O for all k The reflections in O generate Dm

3.3 Main Lemma

may pull a pair of reflections (r, r0) ∈ D2

m to the left most or right most positions of T given gcd(d(r, r0), m) = 1

Proof The case in which T is constant is trivial, so assume otherwise Consider all the pairwise distances of reflections, choose the pair with the smallest positive difference, say (ri, rj) Using Hurwitz moves, we may move any reflection rightward leaving it unchanged Suppose ri is to the left of rj in T , move ri rightward until ri and rj are adjacent We have altered T using Hurwitz moves to form some equivalent but likely different tuple

˜

generated by ri and rj There are two cases, either there exist reflections in ˜T outside of

S, or there do not We discuss both cases separately

If there do not exist reflections in ˜T outside of S, then ˜T generates S, which

gcd(d(ri, rj), m) = 1 Assume there is a reflection rk immediately to the left of the pair (ri, rj) (if there is not move the pair (ri, rj) to the right so that there is) Because gcd(d(ri, rj), m) = 1, we may transform (ri, rj) into (ri0, rj0) so that d(rk, ri0, ) = d(r, r0) with the correct orientation so that rkri0 = rr0 Move the pair (rk, ri0) to the left-most

or right-most positions unchanged and apply Hurwitz moves to transform (rk, ri 0) into (r, r0)

On the other hand, suppose now that there does exist some reflection R in ˜T outside

apart with D ≤ d(ri, rj) Apply Hurwitz moves to (ri, rj) until s or s0 is in the tuple, creating a pair of reflections with distance strictly less than D Continue to reduce D in

4 Proof of the Reflection Main Theorem

We prove the reflection main theorem for the case when T generates the whole group Dm

the reflection main theorem to T as if the group in question is in fact Dk is sufficient We

of T Recall in this case, T may only contain reflections

4.2 Hurwitz Equivalence when Q T = I

We will prove our claim by using Hurwitz moves to transform any T into a particular canonical form In the case where m is odd, this form is (r0, , r0, r1, r1)

The canonical form chosen for even m differs slightly from the odd case When m is even, we will use the following lemma to motivate the choice of canonical form

conjugacy class must be even

Proof Assume for the sake of contradiction the numbers of reflections from each conjugacy class in T are odd Transform T into an equivalent ˜T with all edge-edge reflections to the left and all vertex-vertex reflections to the right We have

T ∼ ˜T = (∆, ∆0) with Y ˜T = I The product of an odd number of edge-edge reflections must be an edge-edge reflection

Q ∆ Q ∆0 = I There do not exist a pair of distinct reflections whose product is I, which

is a contradiction

Suppose T contains 2nv vertex-vertex reflections and 2ne edge-edge reflections Both

nv and ne > 0, else T does not generate Dm T will be transformed into (r0, , r0, r1, , r1) with exactly 2nv r0 reflections and 2ne r1 reflections

We show we may transform T into the canonical forms described above using the following moves

Proof The way we transform T into its canonical from depends on m For m odd, we show that we may transform T into the following

T ∼ (r0, r0, T0)

When m is even and T contains more than two vertex-vertex reflections, we show

T ∼ (r0, r0, T0)

Similarly, when m is even and T contains more than two edge-edge reflections, we show

T ∼ (T0, r1, r1)

In each case, T0 is arbitrary except that we require the entries of T0 to generate Dm Assuming we may apply the transformations above (we will prove that we may in Lemma 4.2), we show how to transform T into the desired canonical form

When m is odd we continue pulling out pairs (r0, r0) left, leaving a tuple of four rightward entries

When m is even, while the number of vertex-vertex reflections is greater than two, we move pairs (r0, r0) leftward and while the number of edge-edge reflections is greater than two, we move pairs (r1, r1) rightward At the end of this process, we are left with a tuple

of length four

In each case, call the remaining tuple of length four τ When m is odd, τ consists of the four right-most reflections of T When m is even, τ may lie in the middle of T as well

We transform τ into the canonical form (r0, r0, r1, r1)

Proof

τ ∼ (r0, r1, rk, rk−1) ∼ (r0, r1, r2, r1) ∼ (r0, r0, r1, r1) The main lemma may be used to fix the first two entries as (r0, r1) The last two entries must then differ by one, since Q τ = I A sequence of σ3’s and σ2’s are then applied to arrive at the canonical form (r0, r0, r1, r1)

In both m odd and m even cases, we have arrived at our described canonical form Lemma 4.2 We may transform T in the ways described by 4.2.2

Proof Suppose T has length greater than 4, by Lemma 3.7 we may transform T into the following

T ∼ (r0, r1, ∆) and continue by moving r1 to the right to obtain

T ∼ (r0, r1, ∆) ∼ (r0, ∆0, r1)

We haveQ(∆0) = r0r1is a rotation through 2πm by Lemma 3.3, which implies the subgroup

T ∼ (r0, ∆0, r1) ∼ (r0, r0, r−1, ∆00, r1)

T has now been reduced to a pair (r0, r0) and the tuple T0 = (r−1, ∆00, r1)

When m is odd, the reflections in T0 must generate Dm because r−1 and r1 ∈ T0 and are distance two apart, which is relatively prime to m

When m is even and T contains more than two vertex-vertex reflections, while the orbit of r−1, r1 only contains edge-edge reflections, it contains all of them We have ∆00 must contain a vertex-vertex reflection, which must be distance one from some edge-edge reflection, all of which are generated Therefore T0 generates Dm and we are done

At this point we have shown that when m is odd or m is even and T contains more than two vertex-vertex reflections, we may transform T into a pair (r0, r0) and T0 such that Q T0 = I and the entries of T0 generate Dm

Below we briefly show without explanation how to remove a pair (r1, r1) to the right leaving some T0 which satisfies the same conditions as above for the case where m is even and T contains more than two edge-edge reflections

T ∼ (∆, r0, r1) ∼ (r0, ∆0, r1) ∼ (r0, ∆00, r2, r1, r1)

This concludes the proof

resolved Before proceeding, we prove the following lemma

Lemma 4.3 Number Theory Lemma

Let m be some odd positive integer Given a fixed k with 0 ≤ k < m, there exist q, q0 such that q + q0 ≡ k (mod m) with gcd(q, m) = gcd(q0, m) = 1

1 pα2

2 pαn

of congruence relations k ≡ bi (mod pαi

i ) for all i ≤ n while q, q0 satisfy the analogous congruence relations ai, a0i respectively

We examine two cases: fix i, if bi 6≡ 1 (mod pi), choose ai = 1 which leaves a0i =

bi − 1 6≡ 0 (mod pi) and hence is relatively prime to pαi

i In the case of bi ≡ 1 (mod pi), choose ai = 2, a0i = bi − 2 6≡ 0 (mod pi) Then ai and a0i are both relatively prime to pαi

i

pαi

i ) for all i Choose q0 = k − q, q0 ≡ a0

i (mod pαi

i ) by construction Since both ai and

a0i are relatively prime to pαi

i for all i, gcd(q, m) = gcd(q0, m) = 1 and q + q0 ≡ k (mod m)

Lemma 4.4 Suppose m is even and 0 ≤ k < m When k is even, the above result still holds, namely there exist q, q0 such that q+q0 ≡ k (mod m) with gcd(q, m) = gcd(q0, m) = 1 Proof Let m = 2α 1pα2

2 pα n

n where α1 > 0 and let k ≡ b1 (mod 2α 1) Fix a1 ≡ 1 (mod

2α 1) and a01 ≡ b1 − 1 (mod 2α 1) so that a1+ a01 ≡ b1 (mod 2α 1) Since k is even, b1− 1

is relatively prime to 2α1 Combining this with the relations discussed in the m odd case

that q + q0 ≡ k (mod m)

that q + q0 ≡ k (mod m) with gcd(q, m) = gcd(q0

2, m) = 1

Proof Since k is odd, we know k ≡ b1 (mod 2α 1) for some odd b1 Using the same method

as in the m odd case, choose ai and a0i for all i > 1 Define ci ≡ 2−1a0i (mod pαi

i ) for all

i > 1 (by 2−1 we mean the multiplicative inverse of two (mod pαi

i ) for each i) Now define

a1 ≡ b1 − 2 (mod 2α 1), c1 ≡ 1 (mod 2α 1), and finally a0i ≡ 2 (mod 2α 1) Since b1 is odd,

b1 − 2 is relatively prime to (2α 1) and by applying the Chinese Remainder Theorem, we

ci congruences, we get q20 relatively prime to m since c1 = 1 which is relatively prime to (2α 1) and ci is relatively prime to pαi

i for all i > 1

As before, we choose canonical forms for each distinct case, first considering the case when

N > 4

When Q T = rk, a reflection, we transform T into a tuple of the form (Λ, rk) When

Q T = r0rj, a rotation, we transform T into a tuple of the form (r0, Λ, rj) In each case,

described in detail below),Q Λ = I, and Λ is in the appropriate canonical form as defined

in 4.2.1 When N = 3, we choose the canonical form to be (rk−1, rk−1, rk) When N = 4, and m is odd we have the canonical form (r0, rj−1, rj−1, rj) When m is even, depending

on the number of elements from each conjugacy class, we either have (r0, rj−1, rj−1, rj) or (r0, rj−2, rj−2, rj)

Proof When N = 3, we would like to transform T into (rk−1, rk−1, rk) Use Lemma 3.7

to fix the right-most entries as (rk−1, rk) andQ T = rk implies the left-most entry is rk−1

T ∼ (rk−1, ∆) ∼ (rk−1, ∆0, rk+1, rk) ∼ (Λ, rk)

We were able to use Lemma 3.7 for the second transformation becauseQ ∆ = rk−1rk is a rotation through 2πm, and therefore ∆ generates Dm We now consider T0

In the case where m is odd, since T0 = (rk−1, ∆0, rk+1), its entries generate Dm because

rk−1 and rk+1 ∈ T0 and are distance two, which is relatively prime to m Since Q T0 = I,

we may transform T0 into its canonical form, from 4.2.1, Λ and this case is complete When m is even, if T contains more than one reflection in k’s conjugacy class, then

T0 generates Dm This is true because we get the entirety of rk−1’s conjugacy class from the pair (rk−1, rk+1) and one of these reflections must be distance one from a reflection in the conjugacy class of rk Again, since Q T0 = I, we may transform T0 into its canonical form, from 4.2.1, Λ and this case is complete

we have that the entries of T0 generate Dm